Simplify; express your answer in exponential form. Assume $n\neq 0, z\neq 0$. $\dfrac{{(n^{2})^{-1}}}{{(n^{4}z^{-2})^{-4}}}$
Answer: To start, try working on the numerator and the denominator independently. In the numerator, we have ${n^{2}}$ to the exponent ${-1}$ . Now ${2 \times -1 = -2}$ , so ${(n^{2})^{-1} = n^{-2}}$ In the denominator, we can use the distributive property of exponents. ${(n^{4}z^{-2})^{-4} = (n^{4})^{-4}(z^{-2})^{-4}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(n^{2})^{-1}}}{{(n^{4}z^{-2})^{-4}}} = \dfrac{{n^{-2}}}{{n^{-16}z^{8}}}$ Break up the equation by variable and simplify. $\dfrac{{n^{-2}}}{{n^{-16}z^{8}}} = \dfrac{{n^{-2}}}{{n^{-16}}} \cdot \dfrac{{1}}{{z^{8}}} = n^{{-2} - {(-16)}} \cdot z^{- {8}} = n^{14}z^{-8}$.